Vikas TU
Last Activity: 7 Years ago
The net attractive drive would be zero.
In the event that we consider a point p and Q precisely inverse to each other on the circle.
Lets expect that the constrain on a little part (straight wire) on dP of the circle is dF1 and same on dQ would be dF2. The course of attractive field is same at both the focuses yet the heading of current would be precisely inverse. Henceforth, the attractive constrain would cross out each other and the net drive on the entire wire would be zero.